A particle P moves in a straight line. The acceleration a m s⁻² of P at time t s is given by a = 6t – 12. The displacement of P from a fixed point O on the line is s m. It is given that s = 5 when t = 1 and s = 1 when t = 3.

Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

1. [*M1] Use of v = ∫adt: [v = 6t²/2 – 12t + C] v = 3t² – 12t + C
2. [*M1] Use of s = ∫vdt: [s = 3t³/3 – 12t²/2 + Ct + D] s = t³ – 6t² + Ct + D
3. [DM1] Substitutes for s and t and solves equations. Dependent on both Ms: [5 = 1 - 6 + C + D => C + D = 10; 1 = 27-54+3C+D => 3C + D = 28 → C = ....., D = .....]
4. [A1] s = t³ – 6t² + 9t +1 or p=9, q=1
5. [M1] Solves v = 0 to find t values: [v = 0, 3t² – 12t + 9 = 0 (t − 1)(t −3) = 0 → t = ..... ]
6. [A1] t=1 or t=3
7. [M1] Attempts to use at least three t intervals: [[∫vdt] from 0 to 1 + [∫vdt] from 1 to 3 + [∫vdt] from 3 to 4]
8. [M1] Evaluates s for one time interval: [For 0 ≤ t ≤ 1, s = (1 − 6 + 9 + 1) − 1 = 4]
9. [A1] Correctly finds all at least two distances (ignoring signs): [0 ≤ t ≤ 1, s = (1 – 6 + 9 + 1) − 1 = 4; 1 ≤ t ≤ 3, s = (27 – 54 + 27 + 1) – 5 = 4 3 ≤ t ≤ 4, s = (64 – 96 + 36 + 1) − 1 = 4 ]
10. [A1] Total distance is 12 m